2015-12-1: Difference between revisions

Revision as of 13:53, 2 September 2023

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Problema:} Fie Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f:[-1,1]\to \mathbb{R}} o funcție crescătoare, derivabilă pe Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [-1,1]} cu Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(0) \neq 0} . Să se arate ca exista cel puțin un punct $c\in (-1,1),c\neq 0$ , cu proprietatea că

2cf(c)+\int _{0}^{c}f(x)\,dx\geq 0

.

$Solutie:\ (Robert\ Rogozsan)$ Daca $\ f(0)\geq 0$ , cum $f$ e crescătoare, vom avea ca $f(t)\geq 0,\forall t\geq 0$ , deci $2tf(t)+\int _{0}^{t}f(x)\,dx\geq 0,\forall t\geq 0$ . Atunci luam $c\in (0,1)$ arbitrar si concluzia este verificata. Analog pentru $\ f(0)\leq 0$ (luam Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} din Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-1,0)} . Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Observatie:} In functie de cum e Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(0)} fata de Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0} , concluzia se verifica pentru Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle orice \ c \in (0,1)} (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-1,0)} ). Nu avem nevoie de faptul ca Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} e derivabila, nici de Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(0) \neq 0} .

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 <math>Solutie:\ (Robert \ Rogozsan)</math>
-Daca <math> \ f(0) \geq 0</math>, cum <math>f</math> e crescătoare, vom avea ca <math>f(t) \geq 0, \forall t \geq 0</math>, deci <math>2tf(t) + \int_{0}^{t} f(x)\, dx \geq 0, \forall t \geq 0</math>. Atunci luam <math>c \in (0,1)</math> arbitrar si concluzia este verificata. Analog pentru <math> \ f(0) \leq 0</math>.
+Daca <math> \ f(0) \geq 0</math>, cum <math>f</math> e crescătoare, vom avea ca <math>f(t) \geq 0, \forall t \geq 0</math>, deci <math>2tf(t) + \int_{0}^{t} f(x)\, dx \geq 0, \forall t \geq 0</math>. Atunci luam <math>c \in (0,1)</math> arbitrar si concluzia este verificata. Analog pentru <math> \ f(0) \leq 0</math> (luam <math>c</math> din <math>(-1,0)</math>.
 <math>Observatie:</math> In functie de cum e <math>f(0)</math> fata de <math>0</math>, concluzia se verifica pentru <math>orice \ c \in (0,1)</math> (<math>(-1,0)</math>). Nu avem nevoie de faptul ca <math>f</math> e derivabila, nici de <math>f'(0) \neq 0</math>.